Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]

Show Tags

23 Mar 2009, 10:17

2

This post received KUDOS

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

62% (02:02) correct
38% (01:03) wrong based on 384 sessions

HideShow timer Statistics

The sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\) of \(n\) integers is such that \(a_k=k\) if \(k\) is odd, and \(a_k=-a_{k-1}\) if \(k\) is even. Is the sum of the terms in the sequence positive?

And as ak=-ak-1 if k is even Hence, a2=-a1=-1 a4=-a3=-3 a6=-a5=-5 etc.

So, if n is even, a1+a2+....+an = a1+(-a1)+a3+(-a3)+....+an-1+(-an-1) = 0 All terms get canceled.

And if n is odd, a1+a2+....+an = a1+(-a1)+a3+(-a3)+....+an-2+(-an-2)+an = an As, ak=k if k is odd, an = n if n is odd Only an remains at the end, which is a positive number equal to n.

Can you explain this statement a little further, I just do not understand how you arrived at that statement Hence, a2=-a1=-1. Regards

Stem says that \(a_k=k\) if \(k\) is odd. So, for \(k=1=odd\) we have that \(a_1=1\).

The sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\) of \(n\) integers is such that \(a_k=k\) if \(k\) is odd, and \(a_k=-a_{k-1}\) if \(k\) is even. Is the sum of the terms in the sequence positive?

We have following sequence: \(a_1=1\); \(a_2=-a_1=-1\); \(a_3=3\); \(a_4=-a_3=-3\); \(a_5=5\); \(a_6=-a_5=-5\); ...

Basically we have a sequence of positive and negative odd integers: 1, -1, 3, -3, 5, -5, 7., -7, 9, -9, ...

Notice than if the number of terms in the sequence (n) is odd then the sum of the terms will be positive, for example if \(n=3\) then \(a_1+a_2+a_3=1+(-1)+3=3\), but if the number of terms in the sequence (n) is even then the sum of the terms will be zero, for example if \(n=4\) then \(a_1+a_2+a_3+a_4=1+(-1)+3+(-3)=0\). Also notice that odd terms are positive and even terms are negative.

(1) \(n\) is odd --> as discussed the sum is positive. Sufficient. (2) \(a_n\) is positive --> n is odd, so the same as above. Sufficient.

Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]

Show Tags

30 Dec 2012, 04:13

Accountant wrote:

The sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\) of \(n\) integers is such that \(a_k=k\) if \(k\) is odd, and \(a_k=-a_{k-1}\) if \(k\) is even. Is the sum of the terms in the sequence positive?

(1) \(n\) is odd. (2) \(a_n\) is positive

In a sequence it always helps to observer a first few terms. Given these definitions : a1=1, a2=-a1 = -1, a3 = 3, a4=-a3 = -3

so it is clear consecutive terms from begining are canceling each other, i.e., 1-1+2-2+3-3 etc

Also, the sum is either positive in which case it is equal to the last odd term or it is zero.

So knowing either the term is odd or that last term was positive helps us know that sum of the terms are positive

Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]

Show Tags

03 Jan 2013, 21:38

2

This post received KUDOS

a1 + (-a1) + a3 + (-a3) + .......

1. n is odd --> last term is an which is positive every other term cancels out 2. an can be +ve only if n is odd which will be the last term same as above hence answer is D

Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]

Show Tags

14 Feb 2013, 04:55

Now this is what i dont understand. They just mention "k" (i guess constant) but k can take any value -1,-3 or +2. SHouldnt the answer be B then because the statement 2 specifically says that an is positive.

Now this is what i dont understand. They just mention "k" (i guess constant) but k can take any value -1,-3 or +2. SHouldnt the answer be B then because the statement 2 specifically says that an is positive.

\(k\) in \(a_k\) is a subscript, meaning that \(a_k\) is \(k_{th}\) term in the given sequence which starts from \(a_1\), thus k must be some positive integer.

Now this is what i dont understand. They just mention "k" (i guess constant) but k can take any value -1,-3 or +2. SHouldnt the answer be B then because the statement 2 specifically says that an is positive.

dont go into complex things. Just visualize the sequence

It can be 2 way

1,-1, 3,-3, 5,-5, 7,-7 ending in negative term The sum will be zero in this case

1,-1, 3,-3, 5,-5, 7 ending in positive term The sum will be the last term of sequence

we have asked is the sum positive ? ----------> is the sequence as per 2nd case ? ----------> is the a(n) odd ? or is the a(n) positive ? both the statements answer these questions so both are sufficient
_________________

Re: The sequence a1, a2,a3,....an of n integers is such that [#permalink]

Show Tags

20 Nov 2013, 02:29

Bunuel wrote:

KevinBrink wrote:

Can you explain this statement a little further, I just do not understand how you arrived at that statement Hence, a2=-a1=-1. Regards

Stem says that \(a_k=k\) if \(k\) is odd. So, for \(k=1=odd\) we have that \(a_1=1\).

The sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\) of \(n\) integers is such that \(a_k=k\) if \(k\) is odd, and \(a_k=-a_{k-1}\) if \(k\) is even. Is the sum of the terms in the sequence positive?

We have following sequence: \(a_1=1\); \(a_2=-a_1=-1\); \(a_3=3\); \(a_4=-a_3=-3\); \(a_5=5\); \(a_6=-a_5=-5\); ...

Basically we have a sequence of positive and negative odd integers: 1, -1, 3, -3, 5, -5, 7., -7, 9, -9, ...

Notice than if the number of terms in the sequence (n) is odd then the sum of the terms will be positive, for example if \(n=3\) then \(a_1+a_2+a_3=1+(-1)+3=3\), but if the number of terms in the sequence (n) is even then the sum of the terms will be zero, for example if \(n=4\) then \(a_1+a_2+a_3+a_4=1+(-1)+3+(-3)=0\). Also notice that odd terms are positive and even terms are negative.

(1) \(n\) is odd --> as discussed the sum is positive. Sufficient. (2) \(a_n\) is positive --> n is odd, so the same as above. Sufficient.

Answer: D.

Hope it's clear.

Thanks. But could a1 = -1? the question does not state that k>0 so I make a big mistake here... Please help to explain. Thanks a lot!

Can you explain this statement a little further, I just do not understand how you arrived at that statement Hence, a2=-a1=-1. Regards

Stem says that \(a_k=k\) if \(k\) is odd. So, for \(k=1=odd\) we have that \(a_1=1\).

The sequence \(a_1\), \(a_2\), \(a_3\), ... \(a_n\) of \(n\) integers is such that \(a_k=k\) if \(k\) is odd, and \(a_k=-a_{k-1}\) if \(k\) is even. Is the sum of the terms in the sequence positive?

We have following sequence: \(a_1=1\); \(a_2=-a_1=-1\); \(a_3=3\); \(a_4=-a_3=-3\); \(a_5=5\); \(a_6=-a_5=-5\); ...

Basically we have a sequence of positive and negative odd integers: 1, -1, 3, -3, 5, -5, 7., -7, 9, -9, ...

Notice than if the number of terms in the sequence (n) is odd then the sum of the terms will be positive, for example if \(n=3\) then \(a_1+a_2+a_3=1+(-1)+3=3\), but if the number of terms in the sequence (n) is even then the sum of the terms will be zero, for example if \(n=4\) then \(a_1+a_2+a_3+a_4=1+(-1)+3+(-3)=0\). Also notice that odd terms are positive and even terms are negative.

(1) \(n\) is odd --> as discussed the sum is positive. Sufficient. (2) \(a_n\) is positive --> n is odd, so the same as above. Sufficient.

Answer: D.

Hope it's clear.

Thanks. But could a1 = -1? the question does not state that k>0 so I make a big mistake here... Please help to explain. Thanks a lot!

We are told that \(a_k=k\) if \(k\) is odd. Now, substitute k=1 and see what you get.
_________________

Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]

Show Tags

10 Dec 2014, 00:05

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: The sequence a1, a2, a3, ..., an of n integers is such that [#permalink]

Show Tags

24 Dec 2015, 10:24

The sequence is 1, -1, 3, -3, 5, -5..

Statement 1 means the number of terms is odd. From the above, we can see that the odd terms are positives. THeir negative counterparts are the immediate next term, which are even numbered terms. So statement 1 is sufficient for concluding that the sum is positive.

Statement 2 suggest the alst term to be positive. In the sequence a positive is followed by the negative of the same value. Sufficient to answer that the sum is positive.

hence, D
_________________

Fais de ta vie un rêve et d'un rêve une réalité

gmatclubot

Re: The sequence a1, a2, a3, ..., an of n integers is such that
[#permalink]
24 Dec 2015, 10:24

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...